Problem: Simplify; express your answer in exponential form. Assume $z\neq 0, y\neq 0$. $\dfrac{{(z^{4})^{-2}}}{{(z^{5}y^{3})^{3}}}$
To start, try working on the numerator and the denominator independently. In the numerator, we have ${z^{4}}$ to the exponent ${-2}$ . Now ${4 \times -2 = -8}$ , so ${(z^{4})^{-2} = z^{-8}}$ In the denominator, we can use the distributive property of exponents. ${(z^{5}y^{3})^{3} = (z^{5})^{3}(y^{3})^{3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(z^{4})^{-2}}}{{(z^{5}y^{3})^{3}}} = \dfrac{{z^{-8}}}{{z^{15}y^{9}}}$ Break up the equation by variable and simplify. $\dfrac{{z^{-8}}}{{z^{15}y^{9}}} = \dfrac{{z^{-8}}}{{z^{15}}} \cdot \dfrac{{1}}{{y^{9}}} = z^{{-8} - {15}} \cdot y^{- {9}} = z^{-23}y^{-9}$.